C++:从静态方法调用继承的静态方法

Question

考虑以下抽象类AbstractEngine:

class AbstractEngine {
    static void init();
    static std::string getName();
};

并考虑以下2个实现者类:

class Engine1 : public AbstractEngine  {   
    static std::string getName();
};

class Engine2 : public AbstractEngine {   
    static std::string getName();
};

并且,init()函数应该getName()根据类的类型调用正确的:

void AbstractEngine::init() {
    std::cout << getName() << std::endl;
}

例如,如果我打电话Engine1::init(),我希望它打电话Engine1::getName()而不是AbstractEngine::getName()

我怎样才能AbstractEngine::init()真正调用正确的实现getName()？

Answer 1

您可以使用CRTP,即创建AbstractEngine模板类,然后在继承时继承AbstractEngine<EngineN>:

template <typename T> 
class AbstractEngine {
public:
    static void init() {
        std::cout << T::getName() << std::endl;  
    }
};

class Engine1 : public AbstractEngine<Engine1>  {   
public:
    static std::string getName() { return "Engine1"; }
};

class Engine2 : public AbstractEngine<Engine2> {  
public:
    static std::string getName() { return "Engine2"; }
};

Live Demo

如果您还需要一些动态多态行为,则可以创建一个通用的非模板基类:

class AbstractEngine {
    //I assume you would have some virtual functions here
};

template <typename T> 
class AbstractEngineImpl : public AbstractEngine {
public:
    static void init() {
        std::cout << T::getName() << std::endl;  
    }
};

class Engine1 : public AbstractEngineImpl<Engine1>  {   
public:
    static std::string getName() { return "Engine1"; }
};

class Engine2 : public AbstractEngineImpl<Engine2> {  
public:
    static std::string getName() { return "Engine2"; }
};